Class 8 Maths Chapter 5: Number Play Multiples, Remainders & Divisibility Shortcuts
Class 8 Maths – Chapter 5: Number Play
(Ganita Prakash Part 1)
Introduction
This chapter explores the fascinating world of numbers — multiples, remainders, parity, and divisibility shortcuts. Students learn how to reason algebraically, identify patterns, and solve problems using logical deduction.
Basic Knowledge Required
• Addition, subtraction, multiplication, division
• Factors and multiples
• Even and odd numbers (parity)
• Remainders in division
• Algebraic expressions
Important Definitions
Multiple
A number obtained by multiplying a given number by an integer.
Factor
A number that divides another number exactly.
Whether a number is even or odd.
Divisibility Rule
A shortcut to check if a number is divisible by another without full division.
Formulas / Concepts Used
• Even ± Even = Even
• Odd ± Odd = Even
• Odd ± Even = Odd
• Sum of digits divisible by 9 → number divisible by 9
• Sum of digits divisible by 3 → number divisible by 3
• Alternating sum of digits divisible by 11 → number divisible by 11
• If a divides M and N → a divides (M+N) and (M−N)
• LCM rule: If A divisible by k and m → A divisible by LCM(k,m)
Solved Examples (Step‑by‑Step)
Example 1: Sum of four consecutive numbers = 34
Let numbers = n, n+1, n+2, n+3
Sum = 4n + 6 = 34
4n = 28 → n = 7
Numbers = 7, 8, 9, 10
Example 2: Five consecutive numbers with greatest = p
Numbers = p, p−1, p−2, p−3, p−4
Example 3: Remainder problem
Find numbers leaving remainder 3 when divided by 5.
General form = 5k + 3
Examples: 3, 8, 13, 18, 23
Example 4: Divisibility by 9
Check 427.
Sum of digits = 4+2+7 = 13 → remainder 4.
So 427 is not divisible by 9.
Example 5: Divisibility by 11
Check 462.
(4+2) − 6 = 0 → divisible by 11.
FIGURE IT OUT — COMPLETE SOLUTIONS
1. Four consecutive numbers sum = 34
Answer: 7, 8, 9, 10
2. Greatest of five consecutive numbers = p
Others: p−1, p−2, p−3, p−4
3. Always / Sometimes / Never
(i) Sum of two even numbers multiple of 3 → Sometimes
(ii) Not divisible by 18 → not divisible by 9 → Always true
(iii) Two not divisible by 6 → sum not divisible by 6 → Sometimes
(iv) Multiple of 6 + multiple of 9 → multiple of 3 → Always true
(v) Multiple of 6 + multiple of 3 → multiple of 9 → Sometimes
4. Numbers leaving remainder 2 with 3 and 4
General form: 12k + 2
Examples: 2, 14, 26, 38
5. Pebble riddle
Answer: 105 pebbles (fits conditions with 3, 2, 5, 7 grouping)
6. Tathagat’s claim
Numbers ≡ 2 (mod 6).
Sum of three such numbers ≡ 6 (mod 6) → divisible by 6.
Claim is true.
7. Remainders with 7
661 ≡ 3, 4779 ≡ 5
(i) 4779+661 ≡ 8 ≡ 1
(ii) 4779−661 ≡ 2
8. Smallest number with remainders (3 mod 4, 2 mod 3, 4 mod 5)
Answer: 59
Conclusion
Number Play builds strong foundations in multiples, remainders, parity, and divisibility. These shortcuts and algebraic reasoning techniques make problem solving efficient and fun.
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